344 from bottom the hoop then position and velocity

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Unformatted text preview: sses move along a (θ + α)so = speed of each + α),˙ (which you could also circle, the − dt2 2 3R 2 obtain by calculating x + y ). The Lagrangian is therefore ˙ ˙ √ d2 2g 2) (θ − αα2 + mgR cos(θ −(α) + cos(θ + α) =− θ − α). (348) L = 2(m/2)R ˙ dt2 R = mR2 α2 + 2mgR cos θ cos α. ˙ The normal coordinates are therefore (341) √ The equation of motion is θ + α = A1 cos(ω1 t + φ1 ), 2g where ω1 = , 3R g cos θ g cos θ 2 2mR α = −2mgR cos θ sin α =⇒ α ≈ − ¨ ¨ α =⇒ √2g ω= . (342) R R θ − α = A2 cos(ω2 t + φ2 ), where ω2 = . (349) R ◦ Note that if θ = 0, then ω = g /R, as it should. And if θ = 90 , then ω = 0, which Adding and subtracting these yields the angles, which when written in vector notamakes sense. tion are 6.44. Oscillating hoop with a pendulum θ 1 1 = B1 cos(ω2 t + φ2 ), (350) The positions of the masses cos(ω1 t + φ1 ) + B2 are 
 α 1 −1 
 
 
 ◦ ◦ (x, y ’s = R − A’s. If B θ = 0, we have θ) (343) where the B)1 are h...
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This note was uploaded on 01/19/2014 for the course PHYSICS 251 taught by Professor Brandenberger during the Fall '13 term at McGill.

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